Question

A man walks 20 km towards south and then 15km towards east. how far he is from his initial position?

Answer 1

Let the man started his journey from pt A and walks 20 km towards pt B ( south ) and then 15 km to pt C ( east )

In the given figure, AB = 20 km and BC = 15 km

By Pythagoras Theorem,

( AB )^2 + ( BC )^2 = ( AC )^2

( 20 km )^2 + ( 15 km )^2 = ( AC )^2

400 km^2 + 225 km^2 = AC^2

625 km^2 = AC^2

25 km = AC

Hence, He is 25 km far away from his initial position.

Answer 2

The given question is A man walks 20 km towards the south and then 15km east. we have to find how far he is from his initial position.

observe the given diagram, the man started walking from point A and he walks towards B 20 km which is south and then 15 km to point C which is east.

According to the figure AB=20 km and BC = 15 km.

we can solve this by using Pythagoras theorem,

The square of the hypotenuse is equal to the sum of squares of two sides.

Here, the hypotenuse is AC and the two sides are AB and BC.

${ab}^{2} + {bc}^{2} = {ac}^{2}$

${20}^{2} + {15}^{2} = {ac}^{2}$

expanding the squares we get, the values as

$400 + 225 = {ac}^{2}$

on adding,

$625 = {ac}^{2}$

taking the square root, we get the final answer as

$\sqrt{625} = ac \\ 25 = ac$

AC=25 km.

The man is 25 km far away from the initial position.

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